次の5つの定積分を計算します。 (1) $\int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{dx}{\sqrt{1-x^2}}$ (2) $\int_{0}^{1} \sqrt{2x-x^2} dx$ (3) $\int_{-\sqrt{3}}^{\sqrt{3}} \frac{dx}{x^2+9}$ (4) $\int_{0}^{1} \frac{dx}{(1+x^2)^2}$ (5) $\int_{0}^{2} \frac{dx}{\sqrt{x^2+4}}$

1. 問題の内容

次の5つの定積分を計算します。

(1) $\int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{dx}{\sqrt{1-x^2}}$

(2) $\int_{0}^{1} \sqrt{2x-x^2} dx$

(3) $\int_{-\sqrt{3}}^{\sqrt{3}} \frac{dx}{x^2+9}$

(4) $\int_{0}^{1} \frac{dx}{(1+x^2)^2}$

(5) $\int_{0}^{2} \frac{dx}{\sqrt{x^2+4}}$

2. 解き方の手順

(1)

$\int \frac{dx}{\sqrt{1-x^2}} = \arcsin(x) + C$

したがって、

$\int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{dx}{\sqrt{1-x^2}} = \arcsin(\frac{1}{2}) - \arcsin(-\frac{1}{2}) = \frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{\pi}{3}$

(2)

$\int_{0}^{1} \sqrt{2x-x^2} dx = \int_{0}^{1} \sqrt{1-(x-1)^2} dx$

$x-1 = \sin\theta$ とおくと、$dx = \cos\theta d\theta$

$x=0$のとき、$\sin\theta=-1$なので、$\theta = -\frac{\pi}{2}$

$x=1$のとき、$\sin\theta=0$なので、$\theta = 0$

$\int_{-\frac{\pi}{2}}^{0} \sqrt{1-\sin^2\theta} \cos\theta d\theta = \int_{-\frac{\pi}{2}}^{0} \cos^2\theta d\theta = \int_{-\frac{\pi}{2}}^{0} \frac{1+\cos(2\theta)}{2} d\theta = [\frac{\theta}{2} + \frac{\sin(2\theta)}{4}]_{-\frac{\pi}{2}}^{0} = (0+0) - (-\frac{\pi}{4}+0) = \frac{\pi}{4}$

(3)

$\int \frac{dx}{x^2+9} = \frac{1}{3}\arctan(\frac{x}{3}) + C$

したがって、

$\int_{-\sqrt{3}}^{\sqrt{3}} \frac{dx}{x^2+9} = \frac{1}{3}\arctan(\frac{\sqrt{3}}{3}) - \frac{1}{3}\arctan(\frac{-\sqrt{3}}{3}) = \frac{1}{3}\frac{\pi}{6} - \frac{1}{3}(-\frac{\pi}{6}) = \frac{\pi}{18} + \frac{\pi}{18} = \frac{\pi}{9}$

(4)

$I = \int_{0}^{1} \frac{dx}{(1+x^2)^2}$

$x = \tan\theta$とおくと、$dx = \sec^2\theta d\theta$

$x=0$のとき、$\theta=0$

$x=1$のとき、$\theta = \frac{\pi}{4}$

$\int_{0}^{\frac{\pi}{4}} \frac{\sec^2\theta d\theta}{(1+\tan^2\theta)^2} = \int_{0}^{\frac{\pi}{4}} \frac{\sec^2\theta d\theta}{(\sec^2\theta)^2} = \int_{0}^{\frac{\pi}{4}} \frac{1}{\sec^2\theta} d\theta = \int_{0}^{\frac{\pi}{4}} \cos^2\theta d\theta = \int_{0}^{\frac{\pi}{4}} \frac{1+\cos(2\theta)}{2} d\theta = [\frac{\theta}{2} + \frac{\sin(2\theta)}{4}]_{0}^{\frac{\pi}{4}} = (\frac{\pi}{8} + \frac{\sin(\frac{\pi}{2})}{4}) - (0+0) = \frac{\pi}{8} + \frac{1}{4}$

(5)

$\int_{0}^{2} \frac{dx}{\sqrt{x^2+4}}$

$x = 2\sinh t$ とおくと、$dx = 2\cosh t dt$

$x=0$のとき、$t=0$

$x=2$のとき、$2 = 2\sinh t$なので、$\sinh t = 1$より、$t=\sinh^{-1} 1 = \ln(1+\sqrt{2})$

$\int_{0}^{\ln(1+\sqrt{2})} \frac{2\cosh t}{\sqrt{4\sinh^2 t + 4}} dt = \int_{0}^{\ln(1+\sqrt{2})} \frac{2\cosh t}{2\cosh t} dt = \int_{0}^{\ln(1+\sqrt{2})} dt = [t]_{0}^{\ln(1+\sqrt{2})} = \ln(1+\sqrt{2})$

3. 最終的な答え

(1) $\frac{\pi}{3}$

(2) $\frac{\pi}{4}$

(3) $\frac{\pi}{9}$

(4) $\frac{\pi}{8} + \frac{1}{4}$

(5) $\ln(1+\sqrt{2})$